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Question
In Δ ABC, if a cos2 `"C"/2 + "c cos"^2 "A"/2 = "3b"/2`, then prove that a, b, c are in A.P.
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Solution
a cos2 `"C"/2 + "c cos"^2 "A"/2 = "3b"/2`
∴ `"a"((1 + "cos C")/2) + "c"((1 + "cos A")/2) = "3b"/2`
∴ `1/2`(a + a cos C + c + c cos A) = `"3b"/2`
∴ a + c + (a cos C + c cos A) = 3b
∴ a + c + b = 3b .....[∵ a cos C + c cos A = b]
∴ a + c = 2b
Hence, a, b, c are in A.P.
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